@@aaryan6088 i am not at all watching it in how to do it i got this recommendation from yt and when i saw this is college level physics i couldnt stop my laugh so i watched it
I figured it out by myself after seeing "conservation of energy" in the beginning of the video. I did not think of using those laws, makes a lot of sense now!
I really enjoy your physics vids! I’m a chemist (not a physicist..) and also a aviation. adventurist. There is a small group of us aviation adventurists working on an aviation loop FIRST attempt (the reason for being vague). I have started working on the physics behind our project but I’m not a physicist. Would you be will to discuss one on one our little physics problem?
This expression only works when there is no friction on the track. If you are trying to make a real vertical loop, I recommend you make the height of your hill at least 3.5 times the radius of the loop.
Great video, but I have a question: If h is equal to 2r then how to calculate the maximum height that the car can reach. In another words the height when the car will fall down
Hello Physics Ninja, I have 2 questions: 1) Why does the gravitational potential energy at the top of the hill have to equal the total energy at the top of the loop? Shouldn't it be less, because at the top of the hill, there is the additional energy from kinetic energy? (m and g would be constant and since we don't know r, for all you know it could be equal to h.) 2) Someone's already asked this before but it wasn't answered properly. At 8:20, you said "we want the contact force to be equal to zero". Why though? When the contact force is equal to zero, the object falls off the track and that's exactly what we DON'T want. (Also you wrote "N does not equal to 0" so was this just bad wording?) Related to this: At 8:08 you said at some point the term in the left is equal to g. Why? Is it because at the very top of the loop, the only acceleration or energy is the downward acceleration is just due to gravity, or g? Thank you.
Can you explain why we have v=sqrt(gR) ? Why do we want the contact force to be equal to zero? Doesn't that mean that the object going around the loop would fall off?
Great questions! if v=sqrt(gR) this means that the Normal force will be 0. If v is bigger than this value there will be a Normal force pointing toward the center of the loop and the car will be able to go around the circle. If Normal force is 0, the track no longer applies a force to the car and basically means it will lose contact with the track. At that point the only force is gravity and yes it would fall off.
@@PhysicsNinja I don't understand how the normal force counteracts gravity in this situation when it's pointing to the same direction as gravity (the center of the loop as you said). It should be helping gravity instead!
@@bennemann The direction of the normal force is ALWAYS perpendicular to the surface. When you at the top of the loop (upside down) the Normal force would point toward the center and when you're at the bottom is would point up.
@@PhysicsNinja I'm talking about when the cart is at the top of the loop. In that situation, both the normal force and gravity point downwards, so what is stopping the cart from falling?
bennemann it is falling! It’s accelerating toward the center of the circle. At this point it’s also moving to the left so it keeps moving in a circle. The net force at the top of the circle must be toward the center of the circle. If it weren’t moving I agree it would fall straight down but it is moving.
If I have 2 marbles of different mass, your equation shows both should make the loop if I give them the height as per the formula. But thats not correct when I do the experiment
Hi @PhysicsNinja, you have proven that regardless of the radius of the loop and height of the hill, when relying only on gravity, the relationship is always the height is 2.5r. My question is, what is the distance between the base of the hill and the entry point of the loop? In other words, the moving object has to travel a certain horizontal distance on the ground once it reaches the bottom of the hill before it starts to incline and go up the loop - for h = 2.5r, what is this distance and how does this formula change if the distance is changed? Also: is there a limit to the length of the object? Imagine an amusement park ride - a train of several carriages and metres long would have a trail behind it as the first car enters the loop, would that not drag and slow it down and prevent it from entering the incline to complete the loop? Thanks.
Hi, thanks for your question. A couple of points. In my solution i didn't consider the effects of friction. Clearly if there is friction, there will be work done by the friction force and some of the kinetic energy of the mass will converted into heat. In order to include friction i would need to know the exact shape of the curve. I could simplify by only including friction over a straight lower segment after the drop and before the loop. This case could be done without too much effort. In the energy equation you would need to account for the energy loss due to friction (work by friction) As for the length of the object, this is difficult to consider. You would have to know how each cart is connected to the other and calculate the forces between the different objects.
So, why does it NOT work to release the object from a position equal to the height of the top of the loop? Why must it be higher? Does the release location have to be higher because then it will provide the object enough speed to make it around the loop? Thanks in advance!
If there were a straight length/segment after the hill and before the loop that experienced friction and a friction coefficient, how could we determine the minimum height (of the hill) and speed (at the end of the segment)?
why is the "normal" force acting "down" in this case? wouldn't the object experience a normal force more similar to a "hugging the road" force that can be seen in a race car going around a steep incline turn?
@@PhysicsNinja That makes sense. But couldn't it be said that if the normal force was applying in the exact opposite direction (as described in ^ vid) that its force would still be perpendicular? I think we're just talking about centrifugal vs centripetal force (in the context of the normal force within the loop). That would also make sense, especially when considering the force that "slams you into your seat" as you come out of the loop (assuming you're in the car)
thanks for the problem, i have aquiestion, i understand how to get the minimum velocity and that happen when normnal force in the top of the loop is 0, but also for the mass to fall from the loop the normalforce has to be 0, can you enlight me? sorryfor my englishiamfromargentina spanish is our natural language
Great and very helpful video. One question: ignoring friction, why must the starting height be higher than the highest point of the loop? Thanks a bunch.
because at the starting point, we consider that the speed is 0 m/s, so the kinetic energy=zero, so the total energy = gravitational potential energy. At the top of the loop, the object has velocity, so it also has kinetic energy. This means that the total E = Eg + Ek, so the Eg at the top of the loop is lower than at the top of the hill, meaning that the height has to be lower (singe Eg=mgh and m and g are constant).
Great question. If there is kinetic friction with the object and the track you will need to know something about the shape the hill because you'll need to calculate the work done by friction over the entire region where there is friction. You still want the Normal to be set to 0 (that's when you lose contact with the track). The 'other' equation I have in my video comes from conservation of energy. If there is friction you'll need to modify this equation to account for the work done by friction.
No you don't have any kenetic energy at top of loop . t top of loop the kenetic is zero , because it's not moving,,but it has a maximum potential energy....
That's not always true it may possess some kinetic energy...the tension of the string would be zero at the topmost point but it will possess some velocity
